Nontrivial example of continuous function from $\mathbb R\to\mathbb R^{\omega}$ with box topology on $\mathbb R^{\omega}$

Question

Nontrivial example of a continuous function from $\mathbb R\to \mathbb R^{\omega}$ with box topology on $\mathbb R^{\omega}$ and usual topology on the domain.

I can give some trivial example like

$f(t)=(t,t,t,....,t,0,0,0,0,0,0,....)$

Like which are continuous. I had proved many nice functions with infinite coordinate need not remain continuous in this setting like

$g(t)=(t,t,t,........)$

So I am interested in finding an example which has non-trivial infinite coordiante and continuous.

I think such function will not exist because we make some setting to arrive at contradiction as in box topology we have much freedom.

Is such a function exist? Please Give me some suggestion.

Any Help will be appreciated

Answer 1

For $n \in \mathbb{N}$ define $$f_n : \mathbb{R} \to \mathbb{R}, f_n(x) = \begin{cases} 0 & x \le n \\ x-n & x \ge n \end{cases} $$ These are non-constant continuous functions. For each $x$ we have $f_n(x) = 0$ for all $n \ge x$.

Now consider $f(x) = (f_1(x),f_2(x), \dots)$. Let $\xi \in \mathbb{R}$ and $U$ be an open neigborhood of $f(\xi)$ in the box topology. We have $\xi \in \prod_n U_n \subset U$ with suitable open neigborhoods $U_n$ of $f_n(\xi)$ in $\mathbb{R}$. The $V_n = f_n^{-1}(U_n)$ are open neighborhoods of $\xi$. Choose a positive integer $m > \xi$. Then $V = \bigcap_{n=1}^m V_n \cap (-\infty,m)$ is an open neigborhhod of $\xi$. For $x \in V$ we have $f_n(x) \in f(V) \subset f(V_n) \subset U_n$ if $n \le m$ and $f_n(x) = 0 = f_n(\xi) \in U_n$ if $n \ge m$ (note $x,\xi < m$). Hence $f(V) \subset U$.

Answer 2

In a suitable sense, there is no nontrivial such function - for every continuous function $f:\mathbb R\to\mathbb R^\omega$ and any $t_0\in\mathbb R$, there is an open interval containing $t_0$ on which $f$ is constant on all but finitely many coordinates. Therefore, every function is "glued" out of functions which are constant on almost all coordinates.

Indeed, suppose this is not the case. For simplicity, let me assume $t_0=0$ and $f(0)=(0,0,\dots)$. If the claimed conclusion doesn't hold, then for any $n\in\mathbb N$ we can find $x_n\in\mathbb R,|x_n|<1/n$ and $j_n\in\mathbb N,j_n>n$ such that the $j_n$-th coordinate of $f(x_n)$, call it $a_n$, is nonzero. Further, we may assume $j_n$ are distinct. Consider now the set $$U=\prod_{k=1}^\infty(-b_k,b_k)$$ where $b_k=|a_n|$ if $k=j_n$ and $b_k=1$ otherwise. Then $U$ is open in the box topology, but $f^{-1}(U)$ is not open - it clearly contains $0$, but it contains none of $x_n$, since the $j_n$-th coordinate of $f(x_n)$ is not in $(-b_{j_n},b_{j_n})=(-|a_n|,|a_n|)$. Therefore $f$ is not continuous.

Note that the assumptions as I have written them are necessary - we can only conclude that locally $f$ is constant on all but finitely many coordinates. This need not be true globally , consider for instance $f_n(t) = \max(0,t-n)$ and $f(t)=(f_1(t),f_2(t),\dots)$. This $f$ is continuous for $\mathbb R^\omega$ in the box topology, but none of its coordinates is constant. (see Paul Frost's other answer for a proof)